Mathematical logic

Question

Suppose the numbers $1$ to $20$ are placed in any order around a circle . Show that the sum of some three consecutive numbers must be atleast $32$.

Answer 1

This can be proved using contradiction,

Suppose sum of three consecutive numbers is not 32.

Then, $x_{1}$ + $x_{2}$ + $x_{3}$ <= 31

          $x_{2}$ + $x_{3}$ + $x_{4}$ <= 31

               :

               :

          $x_{19}$ + $x_{20}$ + $x_{1}$ <= 31

          $x_{20}$ + $x_{1}$ + $x_{2}$ <= 31

Summing all this we get

     3$\sum {x_{i}}$ <= 31 × 20

= 3 × $\frac{20×21}{2}$ <= 31 × 20

[Since $\sum{x_{i}}$ = $x_{1}$ + $x_{2}$ +....+ $x_{20}$ = $\frac{20 ×21}{2}$ ]

= 3 × 10 × 21 <= 31 × 20

= 630 <= 620 (This is contradiction)

This means , the sum of three consecutive numbers is atleast 32.